Termination w.r.t. Q of the following Term Rewriting System could be proven:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(a) → f(b)
g(b) → g(a)
f(x) → g(x)
Q is empty.
↳ QTRS
↳ AAECC Innermost
Q restricted rewrite system:
The TRS R consists of the following rules:
f(a) → f(b)
g(b) → g(a)
f(x) → g(x)
Q is empty.
We have applied [15,7] to switch to innermost. The TRS R 1 is none
The TRS R 2 is
f(a) → f(b)
g(b) → g(a)
f(x) → g(x)
The signature Sigma is {f, g}
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(a) → f(b)
g(b) → g(a)
f(x) → g(x)
The set Q consists of the following terms:
g(b)
f(x0)
Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
G(b) → G(a)
F(x) → G(x)
F(a) → F(b)
The TRS R consists of the following rules:
f(a) → f(b)
g(b) → g(a)
f(x) → g(x)
The set Q consists of the following terms:
g(b)
f(x0)
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
Q DP problem:
The TRS P consists of the following rules:
G(b) → G(a)
F(x) → G(x)
F(a) → F(b)
The TRS R consists of the following rules:
f(a) → f(b)
g(b) → g(a)
f(x) → g(x)
The set Q consists of the following terms:
g(b)
f(x0)
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
G(b) → G(a)
F(x) → G(x)
F(a) → F(b)
The TRS R consists of the following rules:
f(a) → f(b)
g(b) → g(a)
f(x) → g(x)
The set Q consists of the following terms:
g(b)
f(x0)
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 3 less nodes.